How do you differentiate #f(x)=xsqrt(abs(x-1)# using the product rule?
Please see the explanation section below.
We must first analyze the function because it involves the absolute value function.
Thus, we have
and
Combining these results in
Extra remarks
This notation gives us
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To differentiate the function ( f(x) = x \sqrt{\left| x - 1 \right|} ) using the product rule, we treat it as the product of two functions: ( u(x) = x ) and ( v(x) = \sqrt{\left| x - 1 \right|} ). Then, we apply the product rule ( (uv)' = u'v + uv' ), where ( u' ) and ( v' ) represent the derivatives of ( u ) and ( v ) with respect to ( x ), respectively.
First, we find ( u'(x) ), the derivative of ( u(x) ) with respect to ( x ), which is simply ( 1 ) since the derivative of ( x ) with respect to ( x ) is ( 1 ).
Next, we find ( v'(x) ), the derivative of ( v(x) ) with respect to ( x ). To do this, we consider two cases:
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When ( x > 1 ): In this case, ( v(x) = \sqrt{x - 1} ). The derivative ( v'(x) ) can be found using the chain rule, yielding ( \frac{1}{2 \sqrt{x - 1}} ).
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When ( x < 1 ): Here, ( v(x) = \sqrt{1 - x} ). Again, using the chain rule, the derivative ( v'(x) ) is ( -\frac{1}{2 \sqrt{1 - x}} ).
Now, we apply the product rule: [ (uv)' = u'v + uv' ] [ = (1)(\sqrt{\left| x - 1 \right|}) + (x)\left( \frac{1}{2 \sqrt{x - 1}} \right) ] [ = \sqrt{\left| x - 1 \right|} + \frac{x}{2 \sqrt{x - 1}} ]
Thus, the derivative of ( f(x) = x \sqrt{\left| x - 1 \right|} ) using the product rule is: [ f'(x) = \sqrt{\left| x - 1 \right|} + \frac{x}{2 \sqrt{x - 1}} ]
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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